【PAT】【Advanced Level】1064. Complete Binary Search Tree (30)

1064. Complete Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input:

10
1 2 3 4 5 6 7 8 9 0

Sample Output:

6 3 8 1 5 7 9 0 2 4

原题链接：

https://www.patest.cn/contests/pat-a-practise/1064

https://www.nowcoder.com/pat/5/problem/4115

思路：

升序序列就是CBT的中序遍历序列

转化成已知满二叉树的中序遍历序列，求其层序遍历序列

由此进行递归处理

CODE：

#include<iostream>
#include<algorithm>
#include<cstdio>
#define N 1010
typedef struct S
{
	int v;
	int f;
};
S no[N];
using namespace std;

bool cmp1(S a,S b){return a.v<b.v;}
bool cmp2(S a,S b){return (a.f==b.f)?(a.v<b.v):(a.f<b.f);}

void dfs(int l,int r,int fl)
{
	if (l==r)
	{
		no[l].f=fl;
		return; 	
	}
	int i=1;
	int sum=1;
	while (sum<(r-l+1))
	{
		i*=2;
		sum+=i;
	}
	sum-=i;
	i/=2;
	int now;
	if (r-l+1>sum+i)
		now=l+sum/2+i;
	else
		now=l+sum/2+(r-l+1-sum);
	no[now].f=fl;
	if (now>l)dfs(l,now-1,fl+1);
	if (now<r)dfs(now+1,r,fl+1);	
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)	scanf("%d",&no[i].v);
	sort(no+1,no+n+1,cmp1);
	dfs(1,n,0);
	sort(no+1,no+n+1,cmp2);
	for (int i=1;i<n;i++)	cout<<no[i].v<<" ";
	cout<<no[n].v;
	return 0;
}